Wednesday, December 10, 2014

Fun with k-Algebras

Continuing from last time, suppose we have a bilinear map \( f\) from \(V \times W\) to \(L\) where V, W, and L are vector spaces. Then there is a universal propertyfunction \(\Phi \) from \(V \times W\) to \(V \otimes W \) and there is a unique linear map \( g\) from \(V \otimes W\) to \(L\) such that the diagram below commutes:

\(\Phi\)

\( V \times W \)-------> \(V \otimes W\)

\ |

\ |

\ |

\(f\) \ | \( g \)

\ |

\ |

_| \/

\( L \)

The proof is trivial: "f" is used to define a function from the free vector space \( F (V \times W) \) to \( L \) and then we make a descent by modding by the usual equivalence relation of the tensor product to define the map \( g \).

This all looks a bit pedantic, but the point is that any multiplication rule in an algebra \( A \) is a bilinear map from \( A \times A \) to \( A\) and we can now put it in tensor formalism.

In particular consider the algebra \( A \) of matrices over a field \( k \). Matrix multiplication is associative, and we also have a unit of the algebra: the diagonal matrix with the where the elements are the identity of \( k \). This is a prototypical example of what is called a k-algebra.

Can we formalize the associativity and the unit using the tensor product language? Indeed we can and here is the formal definition:

A k-algebra is a k-vector space \( A \) which has a linear map \(m : A\otimes A \rightarrow A\) called the multiplication and a unit \(u: k \rightarrow A \) such that the following diagrams commute:

\( m \otimes 1\)

\( A \otimes A \otimes A \) ----------> \(A \otimes A\)

| |

| |

\( 1 \otimes m\) | | \( m \)

| |

| |

\/ \/

\(A \otimes A\) ----------> \(A \)

\( m\)

and

\(A \otimes A\)

_ _

| | |

\( u\otimes 1\) / | \ \( 1\otimes u\)

/ | \

\( k \otimes A \) | \(m \) \(A \otimes k\)

\ | /

\ | /

_| \/ |_

\( A \)

Please excuse the sloppiness of the diagrams, it is a real pain to draw them.

So what are those commuting diagrams really saying?

The first one states that:

\( m(m (a\otimes b) \otimes c) = m(a \otimes m(b \otimes c)) \)

In other words: associativity of the multiplication: (a b) c = a (b c)

The second one defines the algebra unit:

\( u(1_k ) a = a u(1_k )\)

which means that \( u (1_k) = 1_A \)

So why do we torture ourselves with this fancy pictorial way of representing trivial properties of algebra? Because now we can do a very powerful thing: reverse the direction of all the arrows. What do get when we do that? We get a brand new concept: the coproduct. Stay tuned next time to explore the wondeful properties of this new mathematical concept.